### Isomerization of Methyisonitrile (CH3NC) to Acetonitrile (CH3CN) at Elevated Temperatures#### Chemical Reaction:At elevated temperatures, methyisonitrile (CH3NC) isomerizes to acetonitrile (CH3CN) as represented by the following chemical equation:\[ \text{CH}_3\text{NC (g)} \rightarrow \text{CH}_3\text{CN (g)} \]#### Kinetic Analysis:The reaction is first-order in methyisonitrile. The following graph provides data obtained for the reaction at a temperature of 198.9°C.#### Graph Description:The graph is a plot of the pressure of CH3NC (in arbitrary units) against time (in seconds). The data points form a near-straight line, indicating first-order kinetics. The equation of the line derived from this plot is:\[ y = -5.1 \times 10^{-5} \text{s}^{-1} \cdot x + 5.05 \]- **Y-axis (vertical)**: Represents the pressure of CH3NC, ranging from 3.4 to 5.2 units.- **X-axis (horizontal)**: Represents time in seconds (s), ranging from 0 to 30,000 seconds. Points on the graph indicate that as time progresses, the pressure of CH3NC decreases, suggesting a decrease in its concentration due to the isomerization process.#### Rate Law Determination:Based on the data and the linear equation provided, the rate law for the reaction is to be identified from the following options:1. \( \text{rate} = -5.1 \times 10^{-5} \text{s}^{-1} [\text{CH}_3\text{NC}]^2 \)2. \( \text{rate} = 5.1 \times 10^{-5} \text{s}^{-1} [\text{CH}_3\text{NC}] \)3. \( \text{rate} = 5.1 \times 10^{-5} \text{s}^{-1} [\text{CH}_3\text{NC}]^2 \)4. \( \text{rate} = -5.1 \times 10^{-5} \text{s}^{-1} [\text{CH

Answered: Image /qna-images/answer/5c71a2bd-c62d-4d94-aa9e-82900666f56c.jpg

At elevated temperatures, methyllsonitrile (CH3NC) isomerizes to acetonitrile (CH3CN): CH3NC (g) CH3CN (g) The reaction is first order in methylisonitrile. The attached graph shows data for the reaction obtained at 198.9 °C and the equation of the line is y= -5.1 x 10-5 sx + 5.05. 5.2 5.0 4.8 4.6 4.4 4.2 4.0 3.8 3.6 3.4 0. 10,000 20,000 30,000 Time (s) The rate law for the reaction is Select one: O A. rate = -5.1 x 10-5 s [CH3NC]? O B. rate 5.1 x 10-5 s-1 [CH3NC] O C. rate 5.1 x 10-5 s-1 [CH3NC]2 OD. rate -5.1 x 10-5 s-1 [CH,NC] O E. rate 5.05 s [CH3NC]2 In pressure, CH,NC

At elevated temperatures, methyllsonitrile (CH3NC) isomerizes to acetonitrile (CH3CN): CH3NC (g) CH3CN (g) The reaction is first order in methylisonitrile. The attached graph shows data for the reaction obtained at 198.9 °C and the equation of the line is y= -5.1 x 10-5 sx + 5.05. 5.2 5.0 4.8 4.6 4.4 4.2 4.0 3.8 3.6 3.4 0. 10,000 20,000 30,000 Time (s) The rate law for the reaction is Select one: O A. rate = -5.1 x 10-5 s [CH3NC]? O B. rate 5.1 x 10-5 s-1 [CH3NC] O C. rate 5.1 x 10-5 s-1 [CH3NC]2 OD. rate -5.1 x 10-5 s-1 [CH,NC] O E. rate 5.05 s [CH3NC]2 In pressure, CH,NC

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

See similar textbooks

Similar questions

For the overall gas-phase reaction of the formation of phosgene (Cl₂CO) from carbon monoxide and chlorine, the following mechanism has been proposed: Cl₂ + M2CI+M k_1 CI+CO+MCICO + M k_2 k3 CICO + Cl₂C₁₂CO+CI Derive the expected rate law assuming [M] is large and constant, and using: (a) The equilibrium assumption for the first two steps. (b) Instead using the steady state assumption for any intermediates. (c) Why does the reaction become zeroth order in [CO] in the steady state case? (d) The actual rate law is 3/2 order in [Cl₂] and first order in [CO]. Which mechanism is likely correct? (e) Is the other mechanism possible, based on the observed orders?
Reaction of nitric oxide gas (NO) and chlorine gas (Cl2) produces a very poisonous gas called nitrosyl chloride (NOCl). The experimental data of the reaction is given below. Use the data to calculate the specific rate constant K and overall order of the reaction. Please show full work!
Consider these three reactions as the elementary steps in the mechanism for a chemical reaction.(i) Cl2 (g) + Pt (s) à 2Cl (g) + Pt (s) Ea = 1550 kJ ∆H = – 950 kJ(ii) Cl (g)+ CO (g) + Pt (s) à ClCO (g) + Pt (s) Ea = 2240 kJ ∆H = 575 kJ(iii) Cl (g) + ClCO (g) à Cl2CO (g) Ea = 2350 kJ ∆H = – 825 kJ e. Which reaction intermediate would be considered a catalyst (if any) and why?f. If you were to add 2700kJ of activation energy to the reaction, would you be able to make thereaction reverse itself (i.e. have the products become reactants)? Justify your answer.g. If you were to added a positive catalyst to step (iii) what would the end result be? Justify yourprediction.h. Your friend is looking at your graph and states that she believes that step (ii) is the ratedetermining step. Do you agree with her? Justify your reasoning.
Peroxynitric acid (HOONO2) is an unstable molecule that decomposes to nitric acid and oxygen: 2HOONO2(aq) → 2HNO3(aq) + O2(g)When the concentration of peroxynitic acid is graphed against time, the resulting plot is curved, but if the logarithm of this concentration is plotted, we instead get a straight line. Based on this, which statement is true? a) This decay is a second order in peroxynitric acid. b) The slope of the straight-line graph is the rate constant. c) One needs the concentration of peroxynitric acid to calculate its half-life. d) The rate law appears to be of the form -Δ[HOONO2]/Δt = k[HOONO2].
The rate constant of a chemical reaction increased from 0.100 s−1 to 3.20 s−1 upon raising the temperature from 25.0 ∘C to 55.0 ∘C . Calculate the value of (1T2−1T1) where T1 is the initial temperature and T2 is the final temperature. Express your answer numerically.
to access important values if needed for this question. For the gas phase decomposition of ethyl chloroformate, CICOOC₂H5 C₂H5Cl + CO₂ the rate constant in s¹ has been determined at several temperatures. When Ink is plotted against the reciprocal of the Kelvin temperature, the resulting linear plot has a slope of -1.48 x 104 K and a y-intercept of 24.6. The value of the rate constant for the gas phase decomposition of ethyl chloroformate at 465 K is s-1
Given the following data for the reaction AB, determine the frequency factor, A, of the reaction to three significant figures. k(5¹) 0.730 0.739 T(K) 2.50 102 4.50 x 10² a) -0.287 s1¹ b) 2.05 ¹ Od 0.719 s¹¹ d) 0.287 s e) 0.750 1
Please and thank you ?